Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long INF = 1e15;
long long MOD = 1e18 + 7;
vector<int> ppl[100100];
struct CentDecomp {
  vector<vector<int> > V;
  vector<int> par, sub, ban;
  vector<int> dst[100100][18];
  int N, c;
  CentDecomp(int _N) {
    N = _N;
    sub.resize(N + 1, 0);
    par.resize(N + 1, 0);
    ban.resize(N + 1, -1);
    V.resize(N + 1);
  }
  void build(int u, int p, int l) {
    int n = dfs1(u, p, l);
    int cent = dfs2(u, p, n);
    if (p == -1) p = cent;
    par[cent] = p;
    ban[cent] = l;
    for (auto i : ppl[cent]) dst[cent][l].push_back(i);
    for (auto v : V[cent]) {
      if (ban[v] != -1) continue;
      build(v, cent, l + 1);
    }
  }
  int dfs1(int u, int p, int l) {
    if (l) {
      for (auto i : dst[p][l - 1]) {
        dst[u][l - 1].push_back(i);
      }
      for (auto i : ppl[u]) {
        dst[u][l - 1].push_back(i);
      }
      sort(dst[u][l - 1].begin(), dst[u][l - 1].end());
      while ((int)dst[u][l - 1].size() > 10) dst[u][l - 1].pop_back();
    }
    sub[u] = 1;
    for (auto v : V[u]) {
      if (ban[v] != -1) continue;
      if (v == p) continue;
      sub[u] += dfs1(v, u, l);
    }
    return sub[u];
  }
  int dfs2(int u, int p, int n) {
    for (auto v : V[u]) {
      if (ban[v] != -1) continue;
      if (v != p && sub[v] > n / 2) {
        ++c;
        return dfs2(v, u, n);
      }
    }
    return u;
  }
} * G;
void query(int a, int b, int c) {
  if (G->ban[a] < G->ban[b]) swap(a, b);
  int ta = a, tb = b;
  while (G->ban[ta] > G->ban[tb]) {
    ta = G->par[ta];
  }
  while (ta != tb) {
    ta = G->par[ta];
    tb = G->par[tb];
  }
  int l = G->ban[ta];
  vector<int> V;
  V.clear();
  for (auto it : G->dst[a][l]) V.push_back(it);
  for (auto it : G->dst[b][l]) V.push_back(it);
  sort(V.begin(), V.end());
  V.resize(unique(V.begin(), V.end()) - V.begin());
  cout << min(c, (int)V.size()) << ' ';
  for (int i = 0; i < c && i < (int)V.size(); ++i) cout << V[i] << ' ';
  cout << '\n';
  return;
}
int N, M, Q, a, b, c;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> M >> Q;
  G = new CentDecomp(N);
  for (int i = 1; i < N; ++i) {
    cin >> a >> b;
    G->V[a].push_back(b);
    G->V[b].push_back(a);
  }
  for (int i = 1; i <= M; ++i) {
    cin >> a;
    ppl[a].push_back(i);
  }
  for (int i = 1; i <= N; ++i) {
    sort(ppl[i].begin(), ppl[i].end());
    while ((int)ppl[i].size() > 10) ppl[i].pop_back();
  }
  G->build(1, -1, 0);
  for (int i = 0; i < Q; ++i) {
    cin >> a >> b >> c;
    query(a, b, c);
  }
  return 0;
  for (int i = 1; i <= N; ++i) {
    cout << "For node " << i << '\n';
    for (int it = 0; it < 3; ++it) {
      for (auto itt : G->dst[i][it]) cout << itt << ' ';
      cout << '\n';
    }
  }
}